Consider the family of lines  $\left(\mathrm{x}-\mathrm{y}-6\right)+\mathrm{\lambda }\left(2\mathrm{x}+\mathrm{y}+3\right)=0$ and  $\left(\mathrm{x}+2\mathrm{y}-4\right)+\mathrm{\mu }\left(3\mathrm{x}-2\mathrm{y}-4\right)=0$ . If the lines of these two families intersect at right angles to each other, then the locus of their point of intersection is a circle with radius 

If all the lines of  ${\mathrm{L}}_1+{\mathrm{\lambda L}}_2=0$ are intersecting the lines of ${\mathrm{L}}_3+{\mathrm{\mu L}}_4=0$  in right angle, then the locus of the point of intersection is a circle, whose diameter end points are points of intersections of both family of lines. 
The point of intersection of the lines $\mathrm{x}-\mathrm{y}-6=0,2\mathrm{x}+\mathrm{y}+3=0$  is  $\left(1,-5\right)$ and the point of intersection of the lines  $\mathrm{x}+2\mathrm{y}-4=0,3\mathrm{x}-2\mathrm{y}-4=0$  is  $\left(2,1\right)$
Therefore, the radius of the locus of point of intersection is half of the distance between the points $\left(1,-5\right)$  and   $\left(2,1\right)$
Hence, the radius . $\mathrm{r}=\frac{1}{2}\sqrt{{\left(2-1\right)}^2+{\left(-5-1\right)}^2}=\boxed{3.04}$